1. ## optimization

Find an equation of the line through the point (2
, 1) that cuts off the least area

2. The line must satisfy f(2) = 1, so the slope and y-intercept will vary. We want the minimum area.

$y=mx+b$ with $1=2m+b$,
$b=1-2m$ and $y=mx+1-2m$

Therefore...
$y=m(x-2) + 1$

We want to minimize the area of the 1st quadrant or the area of y-intercept * x-intercept

The x-intercept is when y=0
$0=m(x-2)+1$
$x-int=\frac{-1}{m}+2$

The y-intercept is just b
b = 1-2m

Area must be equal to: $0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{m} + 2 + 2 - 4m)$

Find the derivative and set equal to zero...
$0 = \frac{0.5}{m^2} - 2$
$2m^2 = 0.5$
$m^2 = 0.25$
$m = -0.5, .5$

Slope must equal -0.5 since a positive slope will give an infinite area in the 1st quadrant...

AREA: $0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5))$
$0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5)) = 4$

3. Originally Posted by colby2152
Any more details to this question? Theoretically, you could use the following function that will have an area of zero...

$f(x) = 0, x \neq 2$
$f(x) = 1, x = 2$
But that isn't a line...

-Dan

4. Originally Posted by topsquark
But that isn't a line...

-Dan
Yeah, I re-read the OP's question, and stumbled to no answer at all. I'll look at it tomorrow when I am more awake.

5. Originally Posted by colby2152
The line must satisfy f(2) = 1, so the slope and y-intercept will vary. We want the minimum area OR integral. Considering you do not need need to know calculus for this type of problem, we will avoid integrals for now...

Back to the problem... $y=mx+b$ with $1=2m+b$,
so $b=1-2m$ and $y=mx+1-2m$

Therefore...
$y=m(x-2) + 1$

We want to minimize the area of the 1st quadrant or the area of y-intercept * x-intercept

The x-intercept is when y=0
$0=m(x-2)+1$
$x-int=\frac{-1}{m}+2$

The y-intercept is just m
Good up to this point. The y-intercept is 1 - 2m as you showed above.

So the area will be
$A = \frac{1}{2} \left ( -\frac{1}{m} + 2 \right )(1 - 2m)$

By the way, it changes nothing in the answer, but you should at least deal with what happens in the case of a vertical line or horizontal line, just to be complete.

-Dan

6. Originally Posted by topsquark
Good up to this point. The y-intercept is 1 - 2m as you showed above.

So the area will be
$A = \frac{1}{2} \left ( -\frac{1}{m} + 2 \right )(1 - 2m)$

By the way, it changes nothing in the answer, but you should at least deal with what happens in the case of a vertical line or horizontal line, just to be complete.

-Dan
I was tired... I meant to say that the y-intercept is just b! Thanks for clearing up my work.

7. Okay, I edited my original reply with a full solution. I hope the OP knows Calculus.

8. Originally Posted by colby2152
Okay, I edited my original reply with a full solution. I hope the OP knows Calculus.
Seems to me to be the only way to do it, so I wouldn't worry.

-Dan